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Asked in JEE Main 29th June 1st Shift 2022 · Single spherical surface
R = 15 cm. First surface: (1.5)/(v₁) = (0.5)/(15), so v₁ = 45 cm from the first pole.
This lies 45-30 = 15 cm beyond the second surface: virtual object u₂ = +15 cm.
Second surface (1.5 → 1, R = -15 cm): 1/(v₂)-(1.5)/(15) = (-0.5)/(-15).
1/(v₂) = 1/(10)+1/(30) = 2/(15), v₂ = 7.5 cm.
From the centre: 15+7.5 = 22.5 cm = 225 mm.
→ 225
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